Question

The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7:15. Find the numbers. ​

Answer 1

Answer:

Let the four consecutive numbers in AP be (a−3d),(a−d),(a+d) and (a+3d)

So, according to the question.

a−3d+a−d+a+d+a+3d=32

4a=32

a=32/4

a=8------(1)

Now, (a−3d)(a+3d)/(a−d)(a+d)=7/15

15(a²−9d²)=7(a²−d²)

15a²−135d²=7a²−7d²

15a²−7a²=135d²−7d² 

8a²=128d²

Putting the value of a=8 in above we get.

8(8)²=128d²

128d²=512

d²=512/128

d²=4

d=2

So, the four consecutive numbers are 

8−(3×2)

8−6=2

8−2=6

8+2=10

8+(3×2)

8+6=14

Four consecutive numbers are 2,6,10and14